Wolfram Physics Project: A Discussion with Jim Gates

you want me to start well I think we're live so let's say hello to people I can I can still introduce this it's your show right so for people who are joining us we have been having working sessions talking about our physics project for a little while today we are starting a new and different kind of working session which is talking to outside folk and trying to learn from them about things they've been doing and how they might be relevant or not to our physics project maybe other people finding out things about our physics project too today for our very first of these we're pleased to have Jim Gates who actually I guess we first met like 40 years ago that's correct so Jonathan at Caltech yeah right so they it's some and and then and you were you were already doing supersymmetry is that right at that time that is correct that's where we wrote our book on super space that has sort of become a standard and feel oh that's the that's the thousand and one something with lectures lessons on supersymmetry that's exactly right okay so so so now you've been working out so okay his his some well maybe you want to want to say some things to introduce yourself or maybe we call him sure so oh as Steven mentioned I've known him since 1980 I was a postdoc at the Caltech and there along with a colleague of mine named Warren Siegel and we were working on something called super gravity and this is like a 1980 maybe 2 when that was a very unpopular topic that's right most people that's right most people thought we were kind of crazy in fact I wrote in my teeth first thesis on supersymmetry and when I did that at MIT I was the only person in the entire institution that knew anything about supersymmetry so it's kind of a peculiar situation one of the interesting things as I met John Schwarz and came out to Caltech in 1979 and we started working on a problem and I was there for like a month we didn't solve a problem this problem has ever been solved in the field of supersymmetry no the thing that I'm doing now steven is in fact trying to finish my homework from 40 years ago and figure out why we couldn't solve this one problem and nobody in the world solved it's quite remarkable so you and me both they're working on homework from 40 years ago but so tell me what that what is that problem so let me eat chemists air screen now so I can potential let's go ahead let's stop the video do a screen share and here we go let me come on the side over here I prepared some notes from this as you can see there's a familiar name there on the side is everybody seeing my desktop yes lot lots of lots of points of them files yeah so this is a primer on supersymmetry so let me talk about that first if you look at all the particles in the standard model this is what we see in nature and we have a set of particles that we call force carriers they are all bosons that means they don't obey the Pauli exclusion principle as you well know steam because you started out as a particle physicist ireon right when you were in school to Caltech but all the matter particles are fermions obeying the Pauli exclusion principle so you look at this table and if you'll see most displays of elementary particles they're nice and symmetrical you'll notice this is not and the reason is because I have classified them by whether they are force carriers which is the top row or whether they are objects that the forces are exerted upon and then I classify them do they bain't obey the colleagues who's your best when they don't there's you can see it's a the crazy table would make it like this it's so crazy you might think something is missing and so you might think the world could look like this this is a supersymmetry so what are the current I I don't follow these things in great detail but the efforts to find these supersymmetric partner particles which people were sort of hoping the LHC would be successful at and so on well how far how far away do they have to be to be consistent with current experiments so I think so the LHC which has now ended its second scientific run has put a lower bound on masses like this that is something like something between say 14 and 20 its TTP you know that these things do not have masses less than those numbers so let me ask a question about that I'm sorry to jump into particle conversation the UM so you know back in the day actually when I was at Caltech I did this thing with David Pulitzer where we were looking at you know how the Higgs particle gives mass to particles and well I you know one of things that happened there when you try to have a very high mass particle-like a fermion for example with a the Higgs provided mass the second-order corrections in the effective potential cause the you know cause one to have this instability and the Higgs as a result of the Higgs mechanism how is that avoided for supersymmetric particles so it turns out that in super symmetrical systems you're exactly right a great question but it's super symmetrical systems the vacuums tend to be more stable than in non symmetrical systems so that's part of the there's kind of a built-in damping process so we can push that number much much higher than you would if you were just having nonsense medical systems built so a little it's the Higgs coupling the fact that there's a high Higgs coupling which leads to a big negative term in the effective potential when you have a Fermi on loop there is somehow the fact that that's there is somehow because you're canceling it off from the boson terms that's exactly the mechanism basically looking at vacuum fluctuations or so they damp it out is so you can get away with a lot more if you want to think of it that way but you still have to pair up I mean if you have a you know a 20 TV Fermi on it's kind of got to be paired up with some order 20 T V boson to get it exactly correct when you get to energies let's say close to the Planck scale where essentially all masses will be neglected or negligible but even you don't even have to go at high but you're right the pairing and mass is at high energy skills you expect the spectrum to be equal this is what yeah go ahead I'm curious in in okay let me ask some basic questions here because or you know we're trying to understand in our models for example we're trying to understand how particles emerge in general in our models we understand something about how localized structures can exist in these hyper graphs and so on but we're we haven't yet gotten fermions versus bosons right I mean I asked the following question so you know in you know a defining feature of a fermion is that you know it's a spinor representation it has you know when you sort of rotate by 360 degrees you get a minus sign etc in your setup I mean the fact that we only have fermions and bosons and we don't have you know we have whatever z2 graded algebra is and things we don't the fact that we don't have things with weird Paras statistic right is that obvious why that should be always that's nothing it is not so obvious although so let's just say wait if you make the assumption that only z2 gratings are possible but you what you find is that this is compatible with lots of extraordinary mathematics that you would think is not connected with this at all so for example in the way and we're going to come to that later in this talk I'll talk about yeah and right take care of that see to grading but if you follow the arc of our research say for the last 10 years you'll you'll find things like because this is e2 grading we Oakley are able to get to colossi out manifolds even though this is not string theory we're talking about uh-huh people on surfaces make their appearance a number of just stunning mathematical results result from this z2 grading so we think there's something very special about it but could I as well have I mean if I had you know fermions bosons it's either plus 1 or minus 1 you know could I have some other you know fifth root of of or in fact those are called as you mentioned earlier pair statistics as you probably recall Steve I spent 33 years in Russia Maryland I'm currently at Brown University but at University of Maryland one of the world's experts on pair of statistics or some in some ways the father a physicist by the name of Worley Greenberg I remember that yeah yeah so but so what color was before QCD and color became before it was clear that that you know when you had three you know three up quarks in a what is that a delta plus plus or something particle it wasn't clear how that could possibly work with the exclusion principle and I guess that wasn't exactly right so Warley is one of the people who said maybe there's another kind of statistics at work here and so he's an expert of that and so he actually in while I in the early 2000s was actually proposing experiments that I think some was that were actually done at NIST they did you didn't need colliders do these experiments to look for this parity violation sort of thing and nothing was found but the mathematics he had explored and develop to an extraordinary degree but so just to understand if if we are looking for particles in our you know in our system and you know remember we don't have and we're gonna talk about this at length we don't have you know fixed dimensional space we're dealing with this network which can have an affective dimension that can be even fractional and so on so my question is is the is the fact that in our three plus one dimensional space-time of traditional quantum field theory is the fact that we have fermions and bosons are not some Paris statistics thing something that is in any way related to that dimensionality or is it something where oh you could have you know you know Paris statistics with a permutation group of I don't know however many elements or something well as I said mathematically from the work of wall-e we know that in mathematically there's not anything that forbids that from happening the fact we haven't seen it may have something might well my intuition is I think it may have something to do with Laura's representations but you brought up earlier and also since you brought up networks in the scheme that we're going to talk about which is the set of objects they call it ingress networks are in fact our starting point although they're not kind of networks clearly that you're working on but they are a networks yeah okay so I wanted that that's those they're related to irreducible representations is that correct that's correct okay so I wanted we should we should you you've clearly got a few more slides here I think that one I do okay so let's and it looks like Boris here was was also on our on our chat and was maybe maybe we should just introduce we have at least two other people here we have Jonathan gorod and somebody with you Jim Boris who was actually my former student I taught him quantum field theory okay let me just finish this and then we can start getting ok so I'm gonna talk about the problem that JA that Josh Schwartz invited me to Caltech to solve its this problem if you look at you can see there's some operators I call them D alpha D beta and then there's a factor 2 and this thing called Sigma is basically a gamma matrix and so the question is what in arbitrary space times what are the irreducible solutions to this equation and believe it or not if you don't impose dynamics no one has ever had been able to answer this question and this was the question that in 1979 within the context of something called n equal 4 super yang-mills theory this is the question I tried to answer with Josh Schwartz and we couldn't find an answer after a month and no one has ever found the answers ok so let me understand what this equation is sure so so what are the these D's on the Left covariant derivatives of some kind whether they are they are in fact what often referred to as super charges they are the generators and we're going to talk about this in great detail later but these are the generators of super symmetry transformations they are the things that change bosons at the firm bounce and vice versa ok so there's an anti come--as nature of those guys equaling and what's what's that partial derivative on the right that's a survey derivative that index that you can think it was D mu and more conventional notation so it has a space sugar bar which is a disease zero component and then it has the one which is the X derivative D 2 which is the winder ative and detail which is the Z derivative or in general if you were an arbitrary dimensional space like say in theory were there 11 dimensions then that D that D partial derivative must run over all directions right ok but so let me just understand on the super charges on the left yes these are um super chargers enter a Lagrangian for example as you know it's like a a charge multiplied by the field operator is that right if if you want to talk about some property or formula but the basic idea is let's talk about the SU 3 which is something we all know an su 3 we normally represent the generators right matrices and we represent the states by little little column vectors and you let them use a comma column vectors and then they move the entries of the column vectors around so in that setup of su 3 the mass so called gamma and matrices play the role of the DS here they're the things that cause the motions of the representations that you're talking about these so we got a matrix okay so let's just set some so we're trying to get a matrix representation of this presumably super sort of we really want matrices for the DS because as you well know as I remember your thesis and window places as you will know there are some operators that don't have finite matrix representations so we're not really going to be after a matrix representation I refer to su 3 only as men knology so that people can begin to understand why we look at equations like this ok but so but but roughly women so the representations you're interested in are some infinite dimensional representations is that yes that is correct there there feels over space time and therefore infinite dimensional okay and so just to give me an analogy there I mean if we're looking at the standard Lorentz group and things is there what what I mean and where the you know where that decomposes into and you know include +1 dimensional space-time we've got an Esso 3 or su 2 type thing representing that you know spacial degrees of freedom what is what is there now this thing here the closest thing here are actually the momentum generators as you also well know and when we write P mu in quantum mechanics if you try to write it as on a representation that turns into a partial dirty with respect to X and so these things are more akin to the translation operators in fact these things are called super translations ok ok so I still haven't quite wrap my head around what is going on in this equation actually I'm gonna ask Jonathan since Jonathan didn't say anything yet and Jonathan decoding notations of things Jonathan do you understand what's going on here i I think so but I do have a couple of questions so is it reasonable just to think of that you know these DS as being just odd generators in in a super early algebra as translations that are translations in super algebra okay so they have to specifically have to be translations yes ok ok and these and the representations you're talking about even if they're not finite so finite dimensional matrix representations do they still have to be sort of linear endomorphisms of some kind or can they be somehow rich ok right since you you you can't go to something more general like a like a demodulator yes these ok fair enough um yeah okay fair enough I think I think that makes sense if yeah if they're just so so all so all the odd symmetry generators of a superly algebra can be considered trans super translations is that right the ones that are connected to space-time supersymmetry if you talk to mathematicians they have a whole general class of super out Spurs they have nothing to do with space sign translations but the ones that physicists use the answer to your question is yes ok ok yeah I didn't realize that was a restriction that makes sense ok ok yeah I think I think I see what this means spacetime super sorry not sir good mathematicians have a much broader category of things they call super algebras we're not talking about that broad category they're just talking about things where the D mu on the right hand side equation is a spacetime translation generator and we're going to talk about so to see how these things get realized as the assists even you really want to know what they do on a representation so I'm gonna talk about that shortly so but the point is this equation no one knows the answer to so you know I left the accounting and I guess it was probably the villa's this spring of 79 extraordinarily disappointed that John Schwarz and I had not been able to solve this one problem and this problem has actually never been solved in over 40 years so in the 90s when I was already a tenured full professor and I had an endowed chair you know when you get that kind of freedom you can start doing research even if other people tell you it's impossible solve problems so I decided in the 90s that I was gonna go back and try to understand why we were not able to solve this problem let's talk about sq3 for a moment suppose someone came along and said ok I've heard about these things called quarks and he said ok fine I know about course and then they said I can form bound states in these quarks and you say ok I can find bound States and then they asked well what are the representations of these bound States and then so example it becomes a problem a little bit like adding spin right if we you have single spins their spin up or spin down but if they're bound state of two spins you have three states that are here in this representation and one singlet state or if you had three or if you had three spins then it starts breaking more complicated in a more complicated manner so that's the question we're trying to solve for this algebra so one thing I'd understand about that in the case of su 3 going you know the rep the interpretation of su 3 in terms of quarks doesn't have there are no spatial derivatives involved that it's correct that's and that's one of the fundamentally different things about supersymmetry that's exactly right job yeah okay so so okay so let's go back here so let me understand that I mean so you're saying in if I was just dealing with su 3 and I was just trying to find the multiplier structure of su 3 with with you know random bound quarks and things like this okay actually okay okay actually let me ask you in a more naive question okay so in your super symmetric models is there an interpretation are there spinner like things like the quarks of SU 3 yes thing okay and so so your but but you are trying to find a multiplex structure for this like 11 dimensional whatever it is imperiex bro right but so but that multiplied structure is not one that you think you can interpret in terms of just sort of combinations of fundamental particles is that true that's correct the representations themselves I don't say this have to be mathematically independent functions they won't be dependent on anything they can only depend on space and time they can have different spins but that it's all we allow but so I mean back in the days of SU 3 you know when su 3 was originally being formulated in the early 1960s you know people like murray gell-mann you know thought of it initially in a purely mathematical way with just you know and and then and then it was only later that it became clear that the you know spinor representation of SU 3 or whatever it is could actually be represented you know there were just quarks and you could just say how do the quarks add up and that's how you get protons and so on right so the thing I'm trying to understand here is when you're looking at these representations of these super algebra as I guess yes on the the interpretation of that I mean okay I'm sorry to back up a bit I mean when we're looking at the lorentz at the prank or a group for example for space-time symmetries let's talk about the representations of that and we say oh the various spins of particles come out as the different representations of that yes that we're not imagining that a spin three-halves particle is made from a combination of you know three spin 1/2 particles yep correct there's an independent representation alright so in other words is a different type of thing in the case of SU 3 or you know so called constituent su 3 or whatever it was eventually there was an interpretation in terms of more fundamental particles right in the case of something like the Lorentz group there isn't there's just a you know these are the possible representations so these are the possible spins that can exist and say back Lee right ok so then what you're doing is something is more Alice to that second case added exactly right that's what we're after ok now you want to ask what representation I'm going to show you the simplest representation I'm going let me first fall back up so I'm going to cheat so the problem that John that no one in the world has solved is this problem up here that I'm pointing to with my cursor for arbitrary dimension so you know 11 dimensions for in theory strengthening 10 dimensions but for arbitrary number of space-time dimensions you want to solve what are the representation of this equations if you make no assumptions about dynamics yep no one knows how to solve that after 40 over 40 years no one has ever solved that problem and that's the problem in jonshorter I failed to solve in 1979 when I was visiting Caltech for the first time so in the middle nineties I said okay well if I can't solve this problem that I'm pointing to maybe I can solve a simpler problem namely let's go to a magical world where there's only a temporal dimension hmm and yet you have these operators they can act on so in a one-dimensional world with temporal dimensions all the fields are just functions of town let's call it time they're just all function of time and so let's imagine that you have operators like bid like this for in this lower dimension so I kind of switched notation slightly and since I only have a time derivative then the algebra buff can only look like the thing I'm pointing at it was my cursor now namely the anti commutator of these super charges in one dimensions have the same normalization and one dimensions the only thing that I can't I don't have D Sigma major down again mate all I can have is a chronica Delta and a time derivative so this is the analog of that algebra but in one dimension okay and here it turns out I can solve the representation problem because here's a representation the quantity so I introduced some number of bosons remember with the standard model we saw all the Higgs particle and the force carriers signs are just some number bosons I called them Phi I with a lowercase Latin index and some number of fermions these are like the quarks that we saw in the standard model and so I asked myself on this representation can I realize it and realize that algebra that I wrote and the answer turns out to be yes and now on what you can see is that here I can actually classify the representations if I can classify the matrices that satisfy these conditions okay and the Hat only on the indices of something okay okay so this is a subtlety thank you for asking so I had to introduce some dobrze bosons I'm pointing at those bosons with my cursor after introduces some number of fermions I'm pointing at the firm Yuans with Mike with my cursor with my cursor but the number of bosons so in general so you could ask suppose the number of bosons which is indexed I counts is different from the number of Fermi on which the number K had so that's all we're doing we're using a notation where in principle we're asking could you have unequal numbers of bosons and still have supersymmetry okay okay so that's all this notation story and now if you actually so how do we figure that out well you'll notice that this is a first order differential equation that's a first-order difference equation let me hit this with the second one of these DS so then on this time side is second order but then using this equation I just have addy hitting oh sighs so it becomes first-order and I can I can concatenate I can do that with the indices in different orders and what that does is put a condition on these numerical matrices that I enter claiming I was hunting for representation yep that those two conditions are what you see below right so okay so you can presumably solve those four matrices and exactly so in one dimension the classification problem turns out to be relatively trivial mainly if I can classify all sets of matrices that satisfying these conditions I have found all at least a starting point for all representations and for supersymmetry its look looks like Boris here has a question that's what was so we should let tum that Boris Oscars question thank you so there DS also contain gauge field trade no no no these are just operators let me try to find an example if you'll give me a second let me try to find the example of what I'm talking about here and since we're all on internet you can do this along with me mainly go online to your browser I'm about to pull mine up here oops that's a let me open a different browser window that's just a message from Danielle and you can go to you can go perhaps the spires right you not get to spires force yes okay so I'm gonna do same thing in spire and the way it's amazing Spiez was that it was the preprint system at stanford that existed when I got involved in particle physics back in 1975 exactly it is only involved just a moderately okay little got a command-line interface okay good okay so Boris for your your benefit so here's a sparsely click the sparse link yes am writing to new it's a new spire in the front is no but it's the basic a single function right and then I'll type in find a gauge and if I want find with a deal look you tell yeah well you don't need to do it turns out okay but it knows not in mine case it remembers your name okay so these are papers that I with my with my graduate students and they're here I hope they're online tuned because they're the one if you're get in there talking about Mathematica C but I'm gonna get my graduate students involved okay all right so let's let me go backwards I don't want that one let me just do another fine namely find a J's and just mock so here we have a list of some of my recent publications so let's go back in time to this one so this is a paper I wrote with hazel who presumably is online with us and we're gonna go and just get that preprint and click on the video ever heard of hollow round me that's one of my it's a joke I'll tell you okay you and if you invented that term like that that was a joke so the reason I got this paper is because this is won't profession I'm sorry this whole discussion about had I got dead okay so you okay so this is a discussion about issue three so I'm gonna use it to figure out to explain something's Western symmetries and su3 in the abstract there are these objects that you can call the generators they're just one half of the Pauli matrices if you go to a specific representations but as someone who does mathematics you can just say gee there's just a set of linear operators but those that those are not the actual poly matrices rest you've asked you to they'd be the poly matrices for su dental you know something were different for su 3 they are the lambda matrices are my matrices right alright so you just put in the abstract you say gee they're these operators and they're linear and they satisfy some condition this is the condition where some of these coefficients temper non-vanishing as you can see here and if you pick these coefficients this is the these are the generators of SU 3 so when you saw on the previous slide when I had the B's you should think of the DS as playing the exact same role as the tease in this equation is that is that okay Boris yes now I see it red but then it there is no sort of a dynamics between fermions so to speak yeah there's right because three you don't need fur meows to talk about issue three it's a it's a class Willie algebra it has nothing to do it's look let me just understand sir because the thing I really want to push on later is talking about how you reconstruct given a knowledge of it representations how you reconstruct the group from them and I guess here if I throw you hey I mean those structure constants the F IJ K R M are are characterized those can be computed from the lie algebra yes right um and but they and there's a certain set of possible combinations of those things that represent the possible you know Li groups and so on it's classified by cart on indeed right with nice stinking diagrams and thinkin back ass cuz I was gonna go there an X right okay oh that's nice okay so a way we you've got nice pictures therefore that's what this paper Steve because we got out of a fire okay so let's go back a little bit so here we are back at the matrices right okay and so if you choose to let the TV makers matrices the T's can act on column vectors those who provide the representation and so you can ask well gee if I take these T's and act on these column these column vectors what they do but if you take T 3 and T as you will know these are the ones that define the eigenvalues they're diagonal and so just the relationships that look like this boy i you know i learnt this stuff when I was a kid I haven't I you know it's what T 3 is I so spend t8 is hyper-charged or something or is that did I get it wrong you got it exactly right steam your your memories still good okay good but Tim but that's again then down there you have an anti commutator and those are because it turns out that you since if you choose these T's to be matrices no one says you have to take the difference in the product you can take the sum that it turns out you get a different set of objects that are not the structure constants they're called the D coefficients in old-fashioned strong physics these things used to be used to describe interactions for strongly interacting particles right and in fact if you do the calculations I've actually nicely worked out for you here what are these specific non-vanishing values of the D coefficients so the first thing that in thinking about supersymmetry is always in the back of your mind to have this picture of how classically Altemus work okay so I'm gonna I'm gonna put this away for now because we may come back and use it some more let's just split away can I ask a very naive question sure okay I'm just I'm trying to get a better intuition for the I mean the question that Stephen asked earlier about why you know in kind of in good old-fashioned su3 you don't get space derivatives appearing it correct but you do in supersymmetry but I'm trying to get some sense ok so is this what Jonathan is that it's a definition in other words if you were going to do su 3 you wouldn't stop with just T 1 and T 2 and T 3 you'll go all you get all 8 you get all of the T's if you're billing as right right right ok so it turns out in supersymmetry if you just take the things with the D alphas that I wrote that does not form a closed super algebra it's just like the SU 3 if someone said I want to just take say t1 and t5 right 2 P 1 and T 5 and su 3 and took the commentator's you will find on the other side something new and so then you would take that something no said okay I'll take T 1 T 5 of this new thing check the commutator did start to generate new things every time until you've got to all 80s and at that point you calculate the commutation ah I don't ever get anything new from taking the commutator right and so the point is that it's supersymmetry if you start off with these these they don't close by themselves you have to include the space tiny ridges for the algebra so what would that look like what would the analogous thing look like for Lorentz group does the Lorentz group have that same kind of structure because that is not it does not it so in the Lorentz group let me let me let me think for a second is there easy way let's let's just go out of them online because you know dr. Google is a wonderful thing so let's just put in Lorentz group here it's and see what we can find out there actually I had this diagram on the talk you listen you use Wolfram Alpha we don't we don't this is not this is not what we solve we would solve we can probably tell you things about Lorentz group but we don't we don't solve giving you narrative about okay let me turn the screen over to you and you find me some pictures brother Lawrence okay it's right now say fountain go to Wolfram apple and find me but the article on the Lorentz group which the no no III don't think I mean we're not gonna have anything terribly interesting probably oh I'm so actually go back to the board I think you should go back to what you were finding okay yeah III did it just quickly want to ask so the the you know the this this closure that you need the space time derivative do you have to add in order to get closure of the lie algebra correct does a of the superly algebra sorry do they come in so presumably they don't come in at the level of the the odd generators because those as far as I know only introduced permutations and time derivatives right so they must come in at the level of the bosonic generator that's also it's correct okay okay so in is there a is there a night is there a nice geometrical way you're thinking about like sort of why that happens I mean because they're just the bosonic generators are presumably just derivative operators acting on as I mentioned when we first wrote the algebra okay okay but so so physically you're saying that the engagement of this of the super algebra with with space-time happens through the bosonic degrees of freedom is that what I'm learning from this knot well you actually have to have both kinds of degrees or feeders it in order to super have supersymmetry you must have bosonic degrees of freedom and fermionic degrees for indeed yeah that's one of the defining features of that representation I showed you I should absolutely right yeah okay no no I'm just trying to understand as as you you're saying the lie algebra does not close unless you have these space derivatives that is correct but you're saying that the the fermionic degrees of freedom somehow aren't the ones that have to pull in I mean because they you know the spatial well here to be very naive I mean that the spatial degrees of freedom that the spatial derivatives don't have I mean they're not grassman like things are they yes they're just regular derivatives that have you know that work the same way that that have as opposed to being some kind of fermionic degrees its degrees of freedom correct okay so the so the messages have your one of your supersymmetry you've got to be committed for this as part of that z2 grading it you were asking about young okay wanted to admit the Zija grading at the sort of at the first step in some sense so let's see the the I to derail just a little bit because I'm really curious about I mean what one of the things that well maybe maybe am I think he was you were still explaining the the the representation problem that you're that you're trying to solve but then I I'm at some point I want to jump in and ask a bunch of things about know this is willing discussion so I I don't mind okay okay all right so let me launch into the thing that I guess I'm most curious about right now okay alrighty oh so we have I mean let me here I could perhaps show you fuse to Sherry I could just bring up okay I'm going to put my stuff on the side it go ahead [Music] [Music] okay oops I was sharing there we go okay so I mean you know what we have in our whole setup is you know we have this whole notion of the spatial hypergraph mm-hmm and and the way that you know we imagine particles work in this is by having various kinds of sort of topologically stable structures um that you know can maintain their identity as they propagate through space basically okay so for example to understand the limit of this object when it gets you know the the large-scale amount of this object we understand how to compute dimension we understand how to compute curvature we understand how to make jd6 we understand how to make we just starting to understand all kinds of other things you can make out of combinations of gd6 here okay um which is so far okay so so here's the thing that that we're sort of trying to understand what is spin we're trying to understand that that's something which we expect to emerge from features match the rule example of this if I can pull it up um if I can pull this up um there we go okay so so this is the thing where we are looking at the different ways in which you can do the different ways in which you can match the underlying hypergraph when you do an update and there are different ways you can do that and there are permutations that you can make of the underlying elements and the hypergraph and depending on what rule you choose you get different permutation groups that respond to the the sort of a freedom associated with how the update gets done mm-hmm okay so the speculation is that the way you get local gauge invariance is by essentially a large-scale limit of this kind of ambiguity and the way the updates happen and so one of the questions is how do you take so what we have here is something where we've got at you know it's sort of a local level we're doing these updates on this hyper graph and we have certain updates that are equivalent to other updates so we can define permutations that say which you know which different underlying elements can be exchanged with the return reliance of larger and larger chunks of hypergraph and so we're defining certain certain rewritings are equivalent to certain other ones so there's a certain collection of permutations that leave the rewrite invariant so we've got this collection of permutations and so one of the questions is how does a collector we can measure the dimension of the spatial hypergraph by by looking at the growth rate of JDC balls in it so you start from a node in the spatial hyper graph and you just keep looking at things which are increasingly higher radii away from that that node and if the growth rate of the the geodesic ball the leading term is R to the D then D is zero dimension mm-hmm and but that D doesn't necessarily have to be an integer and so one question is in so I'm skipping around a little bit and I'm going back to something more like the spin case in that case how do we understand the symmetry group associated with so let's say we've got this thing and we know that this network is limiting in some sense to a d-dimensional manifold so we would expect that that thing would somehow have some kind of if it's somehow homogeneous isotropic it would have some kind of s OD symmetry right okay so the question is what is s OD for non integer D how do we think about it what some a can we for example use representation theory to get sort of a a I mean you know what what we're looking at here when we look at the space and we're looking at how does this space limit 2d dimensional space we've got this notion of you know we got this geodesic balls we can build and things like this the you know the thing I'm trying to understand is what about the cemeteries associated with this how do we you know can we for example just as we take limits and there's an unformed thought okay so just as we take limits of this discrete spatial graph to correspond to a manifold can we take limits of something which is determined let's say from permutations or something on this discrete graph and that somehow can map into representations and can we somehow take some limit of collections of representations to be the thing that will correspond to this you know sort of continuous manifold that exists in the lead group so so let me ask you a question Jonathan because you you sort of trigger a thought this is steel incorrect myself Stephen right so you ask what is the s OD and non-fried non-interview dimensions yes that's not a question I've ever thought about before however I do know that mathematicians that have this notion of fractional derivatives and that has to be the appropriate place to start to try to answer the question because you know I would think that one gets take the notion of fractional derivatives and see if you could write they did see if you could write a set of generators for the rotations in terms of these fractional derivative operators and whatever comes out of that that's what I would expect to be the definition of s OD in non integer dimensions okay so I I'm slightly pessimistic about that rule actually we have happen it happens at our company we have one of the sort of world experts on fractional calculus okay and although I have not specifically asked him that question it is my impression that the way the I mean essentially what you're doing in fractional calculus is you know the standard formula integral X to the N X to the n plus 1 you know over n plus one you're you're you're essentially generalizing that by something like an analytic continuation right and and I I think I mean I might be wrong but i i i i don't know i i'm i'm well let let's just think about that for a second i mean in the you're saying that well no i don't get it because because only in when you go to s OD for international space right you're not changing the order of the derivatives no no so maybe i'm wrong but i would think that somehow right trying to write that okay so look we will often write the rotation in generators as matrices however we also know that we can write them as single power x times a partial derivative operator and then look at the algebra of that object it's a that's a vector field right it's a set of network yeah so if you're doing something that's fractional i was thought that gee maybe if I change the derivatives to be fractional derivatives and I look at the algebra of the vector of the set of vector fields that limits to a Saudi then that would be the definition of what I meant by OD for non integer D yeah I understand what you're saying but this is this is why I'm not convinced about that okay in in you know the way of constructing with you know X dot derivative and things like that it's still always even if you're doing you're going to so5 there so 10 whatever you go to you're still just using those same linear derivatives you're not changing right so so I don't think that the you know the fraction ality of the derivative is what's important I think what you need is a fraction ality let's say you have a matrix representation yeah okay what you need is a fractional size matrix in other words instead of an N by n matrix you have to have a you know an N fractional by and half by half matrix I understand yeah right and so a question would be actually not makes me wonder okay so so my thought would be the representations are that they have an independent existence independent of the matrices that happen to form you know that happened to be able to be used to represent the the representation so to speak sure so my thought would be just looking at kind of the skeleton of the representation is there a way if somebody throws you here are bunch of representations what group does this correspond to you know how you know you can answer that question you can say this is the collection this is the stack of representations this corresponds to the following lis group so what what I'm asking is maybe what one can do is to look at a different sequence of imagined representations so to speak or skeletons of representations and say you know what would that's what would that sequence of skeletons of representations look like if the limiting group to which it corresponds is you know so2 and a half yeah yeah yeah I I'm just know what you're saying I have you know look this is not something I ever thought about but what you're there's the calculational scheme that you're describing would be something I would expect you know if you just carry out the calculations you should have an answer these your colleagues or whoever else let's see but okay so so now I'm going to I mean let's let's take okay I'm gonna ask naive questions which maybe would be supersymmetry yes we're gonna get I think super country I'm hoping super centuries relevant to this question we'll see what we'll do the best we can let's go okay question what's that just before we do can can I try and I'm trying to understand Jim's point about fractional derivatives I'm wondering okay it is what you're saying basically because the the way you represent a infinitesimally obviously you take an identity matrix and you plus and you add a times some defeater for some vanishingly small D theta and some a and little so3 right yeah and so it was your point that one way you can think of the fractional dimensional rotation matrix is as being a is as being an infinitesimal rotation but where the D theta is some fractional operator that was Mike because I mean that doesn't seem as crazy as Stephen made it out to seem okay it seems like I sort of that's some sort of tailored local tailor expansion of dimension yeah okay so where does the infinitesimal so if I take this and I look at the thing with look at an infinitesimal I mean that is those are the coefficients of the D theta yeah if I'm not mistaken that will be the coefficients of I'll take an identity matrix so just just to understand how this would work I would take just they in this case I would take a two dimensional two by two identity matrix and I would add this matrix times D theta no well I would say take that operator and let theta the theta that'd be a small a parameter and they keep only linear terms right fine fine fine fine yes but okay so if I keep on linear terms then this thing then that corresponds to this with theta goes to zero okay so fine so what you're saying is what what I should be looking at is that identity matrix plus you know some D theta times this thing here no not really okay there's like so first of all I would back you up and ask you to write the following if I'm in a two dimensional space I can write x time BBY since you're going to be doing the next size partial disrespect to Y minus y times partial derivative of X and that thing is a generator for the rotations on the two-dimensional space with coordinates x and y indeed okay now if I change the definition of the operator D so that it is a fractional derivative yep then I would ask myself so if I calculate this same object where the derivative is a fractional derivative operator what happens on the other side or more generally I would have of course different you know I'd have be different coordinates but this is just the simplest example of what I'm talking about but but if I were to look at if I had a three dimensional I mean normally if I go to a three dimensional rotation I just have a slightly more complicated version of this where your three three copies of that basically right but but but I'm not changing the DS in that case you're not changing the DS no no but the question would be if you looked at an operator D you know D to the one minus alpha for some sufficiently small alpha and then you applied it and you constructed rotation matrix in that way with the rotation matrix you get out act like a rotation matrix and n minus alpha dimensions or n made me the N minus you know right John that's far the question I'm asking if I change yeah of derivative what yes the notion of the generator of the because I like I've never thought about this before that's why this is an interesting conversation but you know so if you have my expert who can answer that that's what I've asked them right I agree that sounds like a plausible line of argument the yeah you may want one possibility is fractional derivatives I mean we we we sort of tried to think about that before and concluded that it were honest thing but wills but okay but just to ask this question about representations because I think it's going to I think it's going to merge with the supersymmetry discussion I might go wrong um but but okay so and and now I'm being naive here because they don't remember the but just the representations of s OD if what what are the if I just asked for the dimensions of the representations of s OD what-what just thought for if I'm remembering correctly for so3 it's just things like you know 2 n plus 1 or something is the successive am i right am i miss remembering that but I can lick those formulas up I think what you're writing R is just the what's called the the what is it the cartons subgroup I think you're writing the newer generation tar cartons so group ok so but what we but what I'm asking is you know is are the irreducible representations of a group you know to what extent can you use them to kind of um ah let's take a look here so they can find here representation theory of SU 2 [Music] hmm um I mean I'm I'm can I mention one way in which I think Jim's idea is actually very interesting which is that [Laughter] Stephen is pushing this direction I would be very curious to know if you take the operator that's on the screen and you you know extended four to three dimensions putting in the Y and the Z you know this is the version this is the XY you would have a Y Z and I I'd be very curious to know what would happen if I took the definition of fractional two ridges and calculated the commutator of the things I'm talking about I've never seen this anyplace in the mathematical literature but it's a valid question I think right right I was gonna mention that the place where I've seen the most similar idea is in category theory so that in category three you know the general notion of the representation is you have you have one object with the notion of a composition and then you define a functor that map's it to a different object with a similar notion of composition and if you're if you then take think about you know a group in its representation the function that map's the group to the representation is some generalized derivative function mm-hm the representation to the to the group back to the group is some generalized integral functor and I'm wondering cuz there is a notion of sort of fractional you know as Steven has kind of made first made me aware of there's this notion of fractional dimensional tensor products and category theory that there is the leading tensor categories and I believe although I'm not quite sure that one way you can construct them is by having an into an integral functor that's effectively a fractional integral and so I'm wondering if if you translate it from the kind of abstract categorical setting back into something concrete in terms of matrices whether it make whether it actually does make sense to have fractional derivative operators and then consider kind of you know infinitesimally cultivations in the dimension sure so like I said I've never seen this or thought about it but expertise on this sort of thing yeah well I just asked whether we could find out two experts on this stuff maybe they'll join nothing in it I like I said I'd really like to know the answer this question now that we yeah so can we just talk for a second about every Super representation I'm just trying to understand because as I understand it one of the things you're trying to do is find representations of these bully algebra that's correct right so so I'm what I'm trying to understand is if we were trying to find out does our model you know does it have some some wonderful superly algebra that pops out of it for example yeah yeah the the question is how would we know because all we see is these things where I mean we're making it you know from the dust up so to speak and we are sort of where the dinkler's also a very similar philosophy that we're sort of building things from the bottom up okay okay that's that sounds very interesting I mean so then maybe we should hear about what the Dinkas are and how they work well if you'll turn the screw back okay go ahead a second you have control for my second Wolfram let's see screen share desktop now because I prepared a bunch of powerpoints oh great okay for us to go through I think I just went through the first one yeah yeah okay so here's the second yci i figured the first one was the only one so i was why would you think that Steven okay okay oh these look this looks this looks very interesting these are interesting but let me take you through it okay oh actually what one person on our live stream Richard is pointing out the fractional dimensional vector spaces of fusion categories and category theory right right that's that's the Deline tensor category is so fusion categories in the particular class of symmetric monoidal categories that corresponds to Deline tensor categories and then the ones on which you can define this fraction do they know the answer to what would happen if we as fractional directors replace the home of the ribs isn't that generators I think they have different X but I'm guessing that their expertise is different but hopefully we're finding our local fractional derivative X okay okay can I also put my five cents for it that would be totally crazy but representation of sson are Wigner matrices yup right I just typed also in Mathematica you have dignity yes magna dignity right so but you're looking at at integer non integer SS that which is a little strange for D matrices but D matrices are also but spin right yes spin spin 1/2 so and here it may well actually yeah you give me an idea okay because we know that for example the three Dre symbols can be represented in terms of hydrogen metric functions right and so that so my question then would be okay what what are the big nerdy matrices in terms of what what is the what are those in terms of mathematical you know other mathematical functions actually Michael you might know the answer there's also some PF cues that PF cues yeah and do you imagine so what then one might guess is that the P and the P FQ has something to do with the with the with the D of the s OD so to speak Lorde's if we were to look at s if we look at so2 as well as known as u1 what will be the analog of the Vigna D matrices and how would they be represented in terms of of hypergeometric functions that's Michael I think this has been done there's this Russian a big clean book form the versions from 2005 I think they have all of this you know the answer to my question that I posed a few moments ago SOI unfortunate event one joins late one does not see earlier question so I don't know two question okay so the question is let's start so I'm trying to get trying to understand this whole fractional dimension thing that Stephen has been talking about in the constructions and so the question is the following suppose that I just considered let's say so3 because it's a very simple group to consider you know I know how to write down vector fields for the three generators of so3 you all know these expressions if I replace the partial derivatives in those expressions by fractional derivatives what do I get out when I try to calculate commutator x' that's the question and if you and I'm not trying to put you a spot if you have never seen this before this is something good way to think about but I'm curious to know the answer to the questions now that I've posed it I have never thought about doing this okay say something I would say a little bit about I know I don't know that Oleg is here also but but can you say a little bit about fractional derivatives and how to think about them in terms of um I mean there's an integral transform that defines these fractional derivatives exactly I think for physicists in my opinion the nicest representation of affection derivative is be a Fourier transform because derivatives become multiplication for your space so normal one might be case by P the plus P squared so you trust basically multiplied by something fractional P to the see half and then define it through the inverse Fourier transform so let's think about that the free a approach thinking about that in terms of what Jim is talking about with these commutator 's what would happen if you just take the Fourier transform of the commutator expression and then try and reconstruct it backwards no I mean the freeze and transform of the products would become a convolution which which is not unexpected because on the indebt sort of differential operators are local non integer orders are non-local so you need is they basically touch the function far away okay so what you're saying is that that what when we're looking at a commutator here X dot d things like that that in integer dimensions that corresponds to a local operation on the function Jim I think this is a highly relevant point here okay well I got you know I'm just look I'm just the I'm just a local mathematical ditch-digger in this game okay well but okay but listen I think I think Michael is making a really good point here so okay so well okay so I can say there's no one knows the answer no no no I think we have a partial answer here I think Mike okay so the point that Michael is making is that first of all think about the the fractional derivative by taking Fourier transform you know multiplying by some some fractional power taking the inverse Fourier transform uh-huh now the point is there is a big difference between taking those Fourier transforms in for the integer dimension for the ordinary derivative and this fractional derivative because basically in any integer dimension that's a local operation in in precision space in position space but in a fractional dimension it is not a local operation in position space physics example that comes to mind that isn't quite affecting the derivative but nearly is the relativistic K square root of P Square plus M Squared because there we know form big nor Newton beak now Newton Vietnam beat us up a door and so on that basically we have to solve in Tikal equations for if we take relativistic energy relations okay okay we don't have a fonts in this but you have part over the answer and we have part of an answer which is good okay oh now you've got any excited well now I've posed a homework problem for your genius assist even yeah right well okay but you've also gotten me excited about I think this Boris's comment got me excited about the continuation addi function yeah okay let's go let's go back I want to know what all these pictures are a dentist so let's look okay so okay so finally back to your dater so here we go yeah right so remember I was claiming that I'm really trying to it's a representation theory problem that I'm after that most of our colleagues have forgotten exist and they don't think it's important where's you know so in the 90s I started thinking about this with one of my students and where think seriesis on that on our chat here has said that they can answer this question it's a really great depth short answer you can use a Dinkas is encoded dimensional precipitants of the fractional dimensional derivative convergences i don't understand all those would i don't understand either but by but my my postdoc said that the fermi like directions of super space are representatives of those fractional coordinates at least for the better at least for one specific value namely one half that's interesting yeah I mean the half would be that half is going to end up being because of its a z2 greatly thing exactly and that's why his comment is also relevant so okay okay so back to the representation theory problem yes please a very bizarre and weird place they asked it by pointing out something that everybody learn when they were in kindergarten mainly that hyper squares hypercubes exists in arbitrary dimensions indeed and so here's the sequence the first hypercube is actually the line segment you move it you know you move it horizontally it generates a square you take the square you lift it up in three dimensions and it forms a cube and this sequence of operation goes on forever so that's the first thing that's actually turns out to be an important mathematical observation where supersymmetry and a ding Chris I mean what's the next stage so let me just concentrate on the two Q here it is so if we're going to make a super symmetric representation first thing that we do is we decorate the cube by putting little you can call them nodes or balls little black and white balls and around the vertices of the square but in an alternating patterns ultimately these white and black balls are going to be fermions and bosons where the white balls are full of bosons and the black balls are fermions the next thing we do is that we color parallel lines the same color this induces from the cube instead of the equivalence classes next we have our strange rule that says put an odd number of dashing x' on every square face in the figure that you're working from and then I can take the object that it results I can do two things to it I can hold this vertex that I'm pointing to a my cursor fixed and let it hang under the influence of gravity and it gives me this or I can hold this side of the object fixed and give the other side of twist in which case it gives me this object the two object you see on the far right hand side are as ingress hold on the hollow those are those are just I'm not very good at deep detangling graphs in my head are those the same graph just rendered differently are they different graphs they are different graphs because in order to get so I one thing I hadn't talked about is that there's well they're they're different if you put enough information on okay so the labels are different but the structure of the graphs is the same or not yes different in a moment I explain why look at so this is how you build a dinkler's essentially you start with hyper cubes and arbitrary dimensions in fact let me just go to that pipe that arbitrary so here's a let me go through the three Q star EQ each vertex gets a white or black ball in alternating manner every parallel line gets the same color on every square face there must be an odd number of - inks so you wait wait a minute wait a minute so this is a coloring so effectively you're saying given any node the node in n dimensions the node has n neighbors for them for the simplest class of a tinker's that's a true statement it is in regular as as my graph there's prints tell me ok so then then in that case from a given node that's white every one of the the N nodes that it is a neighbor of is black is that correct correct is a bipartite graph okay and it's obvious you can do that yes it is obvious for this type of graph it's obvious you can do that yeah what's not so non-trivial is to prove that you can do this in arbitrary numbers of dimensions but if the answer turns out you can write so in other words it's bipartite any n cube and hypercube is some and automatically admits a a decoration of its vertices that's part part bipartite like okay and then and so in terms of this coloring I'm still just trying to understand that so you seem to have every every boson ik node here seems to have salt no it doesn't I'm now confused I don't understand where that where the dashed and and colored lines are coming from okay so let me back up a moment because this is like in the guts of a dangerous so to the left this image we see a cube with its vertices fully decorated mm-hmm now to the right what we have done is every parallel line gets the same color okay so what we're doing is defining a set of equivalence classes on the lines that connect the vertices and there are three classes here red green and black okay fine so every in the D dimensional the things oriented in in dimension n or K out of the out of the D dimensions it's it's green for dimension one it's it's red for dimension 2 etc etc yeah that's correct okay okay you can play this game at arbitrary dimensions that's the opponent right we're going to indeed in D dimensions you have D colors that's correct okay okay so the next step is the dashing rule so let's go back here there knows there's no - shapes mm-hmm the dashing rule says every square face must have an odd number of dashes so you can see that satisfying on the bottom it's satisfied as I get on the top the left and the right okay where each of those faces has one - ings but the front face has to have been three right so I I mean I I don't know why you're putting that constraint but I'm you know I mean I'm gonna explain that in a moment okay so and then front then the final point is that I there's a there there's a kind of an index that happens that has to be assigned to each vertex we call that index height and so one way to get a consistent height here is imagine this is a real object fix this node and let it hang under the influence of gravity that will take you to an object it looks like this mm-hmm and now you can see that the white nodes one here the three in the middle are all of the two different heights and the black nodes are at Heights that are not the same as the white nodes I understand this is hanging in a gravity if I just look at it in terms of graph distance yes but yes you answer your question is yes Stephen okay so it is basically golf distance yeah that's correct okay all right so your version I mean okay that this is why you see you see part of the origin of this discussion was that we had a discussion with an astrologer and some of his graduate students one of whom is your daughter yes I know and so and she said and this is something one always hopes once children are gonna do she said my dad has a theory that might be relevant to this okay she's my agent I have to give her a commission now yeah yeah right okay go keep going yes okay so the object of the far right here isn't the danger oh okay let me go through one more example oh I'm sorry add anchors encode matrices and that's part of the reason why the coloring is important okay so if we decompose the Adinkra into nose that are only connected by black lines or nose that are only connected by red lines or nodes that are only connected by green lines we can use these figures we can map those onto matrices this says the one vertex is connected to the 1 the 2 is the 2 the 3 2 3 before the 4 well that's these that's the 4 by 4 identity matrix women that you're looking at the adjacency matrix here that's correct that's correct you know so you're taking the the subgraph the corresponds to a particular color of edges correct looking the adjacency matrix associated with that sub gulf correct ok and now I can explain the - things so let's do this in red which is on the next line No so if you look in red you can you see that you get these figures this is the first node is connected to the second but the second node is connected back to the first and that's going to be an eye of something in your Jason it's a minus sign you you anticipate a Stephen it's a minus side okay and same thing 3 4 4 3 I don't know if you use this tensor notation but this is a way of writing a 4 by 4 matrix yep okay and then the green turns out to be this so a dealer that's a poly matrix the Sigma is a parametric a signaler Pauli matrix the identities to by the two identity mm-hmm okay so add a curves have the really interesting property that they lead you to Clifford algebras basically ok and so here's the game in four dimension and with the 4q mm-hmm okay essentially that that thing down there reminds me of a hosta Graham that's right that hosta it is in fact that has the diagram okay okay so I now introduce you to the basics of a tinker's mainly that start with the hypercube do this adjusting the vertices decorating I'm sorry decorating universities decorating the links according to these rules and then the final rule is that you want to as you said you want to look at the graph distance between Percy's and you put some rules there even you enforce them all you get a dangerous and it turns out that when you do that you find solutions to that one dimensional version of the problem but I couldn't solve with with huh okay so I'm still trying to ingest this for a moment so you're saying that so first of all there's a unique bipartite there's a unique way up to black-white exchange to to color the original hypergraph the hypercube I would have said there are equivalence classes doing that because you know color is just some kind of placeholder in our mind for the definition of equivalence Basson's you could use any colors right no no I understand it but I'm saying that in terms of which node gets the fact that what is not immediately obvious to me is that there's a single pattern of coloring the nodes black and white and ah not the Disco's actually this goes back to something I said earlier that it's not obvious that you can do this it's the dashing look with a little bit of work you one can prove that the alternating black white is really simple for hypercubes what's not so obvious is is that you can put in the rule that every square face has to have an odd number of dashing that's the one that you have to work a little bit for mm-hmm okay okay all right okay so as I said so so one so as you mentioned even these are basically adjacency matrices and here you would have to have four different adjacency matrices one associated with each color mm-hm and now let me go back to the problem that I was first showing you where they put it so I'm sorry that then of these things in terms of those retro the representations all vary this superly algebra stuff that is correct that's exactly right okay yes by that means that I just showed you which is starting to just starts from a bunch of geometrical dressings of graphs if I look at the adjacency graphs I find solutions to the set of equations you see at the bottom but so what is the family I mean so given a given a lea group there's a presumably infinite family of irreducible representations of that direct also so so now how do you get is it is it what sequence of Adinkra is give you that sequence of irreducible representations we don't know the answer that question what we can tell you is that we know how to find what because the construct that I just showed you leads to the minimal for any given color or at least the minimal representation for any fixed number of colors but we don't get a theorem or a way to classify the higher members of these irreducible families okay so just so I understand if I want to make okay being naive if I if I look at like a rotation group some some very simple thing was simple that that could be defined by a simple set of commentators and things like that yes is are there a dinkler's that correspond to the rotation group yeah well in a sense yes because if you put them up I'm sorry let me take a listen I can remember where to find this let me let me show you one other thing let me say it this way Stephen suppose I put minus signs here mm-hmm then you have objects that effectively are the orbital angular moment I'm sorry are the spinor representations of orbital angular momentum yep and so yes they sort of nor I miss construct naturally leads you to double covers well lead you to Clifford outdoors and ultimately to rotation algebras always so just as a naive level I mean if if if you took your dinkar like construction I mean so there were all these things I mean I don't really remember them very well but in fact you know with young tableau and things like this for know you know constructing the irreducible representations yes sir and so is your identify as 'm I mean can you map some of those things into your Adinkra formalism is there a way like for example instead of taking black and white you know nodes you just have one color of nodes and different rules and so on do you so Steven the question is a good one we don't know the answer because remember I specifically set out to study supersymmetry and there are only two classes of particles and super symmetries namely fermions and bosons so I must look at graphs that have been apart part time to represent that okay but so maybe you should explain because I'd done the relationship to comfort algebras so oh okay sure hang on a second I'm gonna have to but so so I mean what I'm be curious about from Europe which is sort of perhaps relevant both to your kind of thing and I kind of thing is you know is there a nice graph theoretic rep you know interpretation of representations of arbitrarily groups I would say the closest that I know to that are the weight spaces that you know just as I do those are sort of graphs yeah I mean I I don't yes I mean I my knowledge of them is pretty pretty decayed at this point but but I mean there's also a thing that I don't really understand very well maybe you've studied my friend predix fatahna which has his whole sometime last year of the year two ago right I've known him even longer than I've known you okay the but I don't understand I mean he it took him like thirty years or something to write his book about bird tracks and League right yeah I have a copy of that book in my computer that do you know whether there's a relationship between what he's doing and what you're doing well I don't know but you know I went part of the reason I was I would met him was to talk about that very question we you walked away not knowing the answer okay fair enough ah okay this doesn't show it so well but the point is that our Ellen armed our Ellen our major see so start with a Clifford algebra in a Euclidean space that's what number twelve tells us yep if you start with the Clifford algebra in the Euclidean space then you can always take one of those different elements and use it to find projection operation with respect to the remaining hit mm-hmm then you can once you have these projection operators you can act on the remaining in elements of the Clifford algebra and that generates these things called L and R that we see here but these nails and ours are exactly the things that I get from the graphical construction so this is only let me just try to understand this so I mean the Clifford algebra is normally you would have what two to the N dimensional I mean you know there's a the let's let's just the what's the relationship so first of all a d-dimensional Adinkra what is the D in the D dimensional Adinkra what does that de correspond between the D dimensional in a D course actually corresponds to the number of for bosons or fermions in a super symmetric representation so as I look at like the super symmetric analog of of some particular well what's the simplest what's the simplest one of these some what's the simplest example where there's a of a of a supersymmetric come the sent was on typical one I'm about to put up in front of you it's this maybe this is the this is the simplest non-trivial Adinkra yep and I've decomposed into its different colors which leads to these adjacency matrices and you use these adjacency matrices in the formula that I showed you for the realization of the supercharged which was that I said you have to do what's on this an edge namely when you say the supercharged axon a boson it gives you this L matrix when the supercharged X is on the Fermi on it gives you this matings but times a space show here this space only has one direction a space so down down below your anti commutator is what's the what's the interpretation of those anti commutator Zanda graph theoretic terms that's oh by the way that's what all those plus or minus signs are about by the way it's the - eyes that enforce those conditions that the dashing no I get it I get it but I'm just I guess I'm I'm asking if you take you know L l dot R effectively you've got a product of two adjacency matrices all right let's correspond to the two-step you know that that's you know the product of those adjacency matrices is like you go first on an L edge and then on an RA that's correct okay so if you take the trip in one way if I can call the motion to trip you take it you actually take applicat you pick color one color - it takes you from one place in a representation representation space to another place now I'll do the same thing but choosing the colors of the links in the opposite order when you add those two things together that's what these two equations on the pocket so so to what extent is this like coverture in that space god it is almost like curvature except that the sign is wrong okay but but now in the case where you're not dealing with the Souper example the sign probably is you know if you didn't have the super business you would have commutations not only anti commutator said it's correct so in what sense then is the I mean is this thing the analog of them so you're saying in this Adinkra space so to speak these the the presence of these commutator x' it's it's say there's a certain amount of curvature but not well I'm not sure that it's saying this maybe it's saying that zero curvature I'm not sure the closest analog would be zero curvature but non-trivial portion right yeah in fact it's interesting thing in our actually there's a question for Jonathan here do we end up getting torsion in any of our metrics or do we do we are we forcing it not to have torsion yes the construction of our space-time metric assumes torsion freeness but but that doesn't necessarily hold and yeah this is that's a really interesting point that it yeah an interpretation is that curvature and supersymmetry becomes effectively torsion I hadn't thought about it that way before yeah what's always what are you saying you're saying curvature in but but these are I mean these are small discreet spaces but one could imagine in a bigger Adinkra so to speak that the thing would be closer to a continuum limit of curvature or is that is that not correct I mean in other words if I were to take instead of a one one move L&R matrix I was to take a multi move you know if I was to take multiple moves through this Adinkra graph what would that correspond to I mean that that's I'm just trying to know no you're asking the right question Jonathan I'm sorry Stephen and that I don't have good we've actually done some best part of what this whole notion about Halle Roman you asked me about this funny word before is precisely this point is that if I try to take the conventional definition of how I measure curvature I would want - science between subjects not a plus sides exactly and when you do that you do get matrices out the other side that roughly speaking are kind of like curvature so that's what the hollow roaming was all about I see I see so okay but coming back then okay so back back to your Dean Chris so you're you're saying I mean the problem you're trying to solve and the thing you were talking about with N equals 11 super gravity and all this time which by the way okay so the colors are indices of generators is that right that's correct okay and and and so and each vertex do you have separate vertices for the time and space derivatives or does the white vertex or black vertex also contain Titan space derivatives yes yes sorry I should we should we hit ID ourselves yes the nodes are in fact functions of space and time okay but but both the fermionic and boson one so just know both okay interesting interesting I'm very glad that Jonathan and I can be mixed up because Jonathan is very much younger than I am and so my students who are on this call with us will tell you that I often mention to people I have a single active brains over me oh yeah it's I don't know it then it must be doing very well you do my claim is it dukes us because see if the input output rates are fantastic uh-huh then but but so actually we had a number of questions here on the on the live stream so okay um in here does the coloring correspond to grading in the coffin algebra no unfortunately not the grading comes from the dashing rule okay oh really yeah okay because I was gonna ask when Steven mentioned before about this you know these things looking like hussa diagrams right so you know it's kind of an accident here but okay so in happen in the higher dimensional cases is that right not necessarily but when you get this partial order does that not give you I mean does that not give you some notion of grading that you know that the height of a vertex or a might completely that is when grading that's what business call engineering dimension it says that's a Z in grading right right okay yes that's what I was trying okay I understand but the permisos grading is the coloring white or black of the nodes yes it's gonna it's confusing that we have to write different notions of grading going yeah yeah then there are - there's the integral grading which is the height of a node in diagrams and then there's the weather notice black or white which corresponds to Fermi on boson so it's as each in there that doesn't make sense so the thing that you've recently solved is is to be able to deal with this 32 my students who are on the call but I think hazel are you are you here hazel yes thank you yeah I guess also here boundary okay so these are my two graduate students hello that and for just over a year we've been making breakthroughs that allow us to so so far in the discussion I'm only concentrated on a decrease in with one dimension because it turns out they are fast they may have a ton of mathematics and I think my colleague Kevin eegah is on is on is also on a stream with us and chem there's actually written a book for mathematicians well it's not not quite finished but it's now at publishers there's a book on math with for mathematicians about exactly what our Adinkra I would recommend that when you know if it ever goes to get published that you and Jonathan and your folks there might be interested in looking at and it gets some but just out of curiosity if somebody can say what are their mathematical buzz terms that connect to things like these Deline categories and so on that come on Kevin if you're online can you speak to us okay well okay no matter no but here's the use of my of the folks that I regularly collaborate with he is the guy that I would put the question to it fair enough so but okay so you were about to explain the the adventure of N equals 11 and so well yes and for that purpose so in one dimension I showed you the rules for constructing these things in 11 dimensions or after anything graded in three dimensions the rules actually become rather different and more interesting and so for the last year a little over a year I and my graduate students young ray and hazel hazel was is my student she's the hazel you're in your third year fourth year oh yeah poor dear poor hazel and young ray you're in your third year okay so these are my fourth year and third year students for the last year so we've actually been figuring out how do you construct an inker is it there is more than one spacetime direction and the answer turns out to be a rather interesting adventure and so I'd like to actually if they're not too shy I'd like to ask one of them to sort of take us through this part of the story and I'll quit sharing here and maybe one of them can share speed or something who wants to show terrific okay so this is a Nadine Kura in ten dimensions the sort of the most basic thinker that you can construct so let me try to understand that so this is this is a essentially just a summary of the Adinkra hanging from its from one node absolutely yeah and so that that's telling you that the oh there we go okay so that that's telling you the multiplicity those numbers of the multiplicity of nodes of that type what are the two different when you see two different nodes or three different node types at a given level what do those correspond to so one of my students are supposed to answer that yes hi I'm young ray I'm I'm James graduate student I'm in my third year PhD first I want to start with this screenshot okay this equation this equation is one of the main results in our latest the paper publishing Jacob and this paper is about the super field component accumulation is in 10 G super space so now we are working on 10 dimensions and in 10 d super space since the my around a while spinner has 60 real components so if we do a sit expression for a general super field we can get zeros order onto your 16th order that's why what expansion I just did there was one word I didn't hear that I said expansion say there's a grassman coronet in super space oh ok theta X match ok so I need to try to understand that so you're saying that you're taking actually that part I don't quite understand okay so in super space we extend our space-time so we not only have the space-time we coordinate we also have the grassman corny to represent some fermionic properties so this is a scalar super field yes the V yeah and the level 0 to level 16 corresponds to a picture for Steven and Jonathan and his colleagues a picture of the super field itself as opposed to just talking about this representation you know 5 V of X of data do you have a similar convenient by the way first people I mean let me just respond to a couple of things on our lives you're asking about all sorts of different things here ok so first of all somebody was asking about torsion and torsion it's supposed to be 0 and general relativity yes ordinary general authority has zero torsion but you're correct Stephen but super gravity never has zero torsion ah ok in the in the metric of super gravity yes it's not the metric the torsion shows up because of the connection you don't use a minimal connection in super gravity you don't use the Christoffel symbols to define the connection is super gravity oh and so so I see so the the torsion is showing up because of a difference the usually the Christoffel symbols have a certain symmetry which they don't have in super gravity that's correct because super gravity you let gravity know generate the torsion so let me understand that so you're taking a covariant derivative and normally that would just extrude Christoffel symbols so to speak that's correct and and what you're saying is that the and those Christoffel symbols usually would be you could interpret as being parts of the gravitational field it was saying in addition you have a gravity no field for super gravity yes and that this and that the gravity no field so what does it mean that the that there's nonzero torsion what do what does that mean it means that metric okay so the role of the the role of the connections will tell your father with the define differentiation on the manifold yep but that means that that means that there are different definitions of differentiation depending on your choice of the of the of the connection and so you can make a choice where's Makino which is different from the choice and when it's not dependent exactly and super gravity always makes a choice to have it so asking on the live stream how does torsion show up in physics what are the physical implications of torsion I think it's that the it's some well maybe maybe somebody else can explain I mean the the the immediate thing is the metric tensor Jimmy nu is not symmetrical and mu and nu well that's one way to say it I the way I sort of so if someone gave you a gravitation covariant derivative since for the people on this discussion at no point have we introduced the gravitational covariant so let's be clear about that right yeah for those are not gravitation Cove area those are operators who are analogs of the gamma matrices I tried to make that clear mm-hmm but in general if someone gave me a gravitationally covariant derivative I could calculate its commutator yep if the Christoffel connection is used I know that when I do that calculation all I get curvature yes however if you use a connection that's different from the Christoffel connection and carry out that same calculation you will find that the right-hand side of the equation includes a term that is linear in the derivative operator that you're using the coefficient of that is the torsion okay so you're saying normally when I take cover in derivatives and I take the commutator I'll just get a bunch of Riemann tensor x' and things around the other side correct that's just the curvature and you're saying in the case where there is torsion I get out something which is linearly proportional to the original derivative operator that's correct and so what is the physical interpretation so I mean if you have a torsion I'm trying to remember there are generalizations of Einstein's equations oh ok so if you think of it in terms of geometry let's do that first when you calculate the commutator what you're doing is calculating an infinitesimal you're calculating an infinitesimal trip around around a quadrilateral in two different ways and then asking what's the difference yep right and the difference turns out to be if you had a vector it gets rotated yep and those rotation coefficients of the Riemann curvature tensor mm-hmm if there's torsion in addition to that rotation the vector field is also subjected to an additional translation and the torsion is a measure of an additional translation oh I see and so so in other generalizations of I'm just this is general knowledge question but I mean although I have a vague recollection that there are generalizations of einsteinium gravity that have some that have torsion and then give different properties with respect to I don't know black holes or whatever else right right my understanding is the you get this additional tensor structure that occurs basically because you get a Li bracket of vector fields appearing because because of this additional vector transformation that you mentioned yeah and does anybody is there any serious theory that has torsion oh there are lots of discussions of torsions and cosmology of national physics it's it's a perennial discussion ok fair enough fair enough okay let's go back I'm sorry yeah angry was was they okay so she's showing here this is an expansion of the verb what is the equation four point one is the general scalar super field in ten Diego where surface phase function in small theta of the value of V that right that's correct yes okay and so so these Thetas are grassman variables right right yep okay so so in principle when you get to the what why isn't there a theta theta theta beta theta alpha that is not the same as theta alpha theta minus because in 10 D we have the if you consider the clear for algebra mm-hmm and we have the Maya run a while condition for the spinner that that means the my runaway laner has the 16 real components now like in one dimension we just have one theta but in ten dimension we have 16 different Sita is the question I'm asking is maybe I'm being silly here but but given that the Thetas do not commute these are non commuting quantities correct anti-community I see so for us specify to say that it cannot be squared no I get that right okay fine I understand all right so that okay all right so this is an expansion there so now that's why we can't only have the level 16 we just learn to sit at 16 because the wedge product I don't actually understand why there are wedge products inside these different levels other such products in the sense of P foam wedge product yes they are is that the grassland variable itself is anti commuting it's not it's not that it's non commuting its anti commute yeah I get it and because of its and communities whenever you write two of them they're automatically forming a wedge product okay or if you three it would be the three words you know so it is the anti community of the grass and coordinate that induces the weights product and so so wait a minute so these different levels what what does it mean so you've got it okay the sixteen is like those are the components of the original original spinner type thing and then you're forming of these different levels okay I'm a little bit lost to what what the 16 is the okay so now we are working on the Norns group right mmm-hmm the 16 is the spinor representation for the Lorentz group as a one common line because we're in 10 D yep yep and now in equation 4.2 we're using the wedge product to count the total Tigre freedom at each level yes I understand but these levels correspond I mean should these levels be thought of as I mean okay naively I got a bunch of spinners you know maybe they're you know maybe physically you know make it up electrons or something and now I could imagine having combined object quarks let's say you know combined objects that have let's say you know five quarks altogether and I have some something which is the you know I'm asking what are the what what are the what are the combined sort of generalized spins that you can get is that how I should think about these different levels or what how should I think about these okay so we can't go back to equation for power mm-hmm so the first the first time is zeroes we rename in the level 0 because there is 0 theta yes so the degree of freedom of the Phi 0 is just 1 yep yep okay look at the second term oh okay I get it so this is just a representation of this expansion this is just saying in by level n you're just saying the the nth derivative is this thing that has a bunch of has n anti commuting Thetas each and each theta has 16 components okay okay I think I understand so far okay so yeah we can calculate so just like the equation 4.1 the level one we have the fine alpha and level two we have five alpha beta but this is just most a general form yes yes we need to find out that you read useful documentation for like the Phi Alpha Beta yep yep so let me let me go back to the mathematical file alright so by using some group theory tool we got finally this is the our final result okay so this is saying and actually yes at each level these numbers represent the dimensionalities of the component views sitting in this level and they are also the dimensions of the irreducible representations of a saw one common life of the spinor representations okay hold on hold on so s o1 nine nine eleven whatever is okay first naive question it they're just one spinor representation of soso1979 which is often mentioned 1660s yeah like representations which means for you have wave in it so the spinner is like and again I'm not familiar and I'm not as used to working in okay so it's like the spinner is like the spinner half creature in in so let's see this you're saying this thing has 16 components are saying it's it's what is the dimension of a coffered algebra for D dimensional space it's it's 2 to the D over 2 or something is that right something like this true right so so then your 16 here is like a spin 1/2 particle is that right yes yes and so then your your your further ones up here is is it the case that your blue ones are the bosonic things like spin one and then the next ok so then so then okay naive question okay so so if I were to look at the corresponding thing in three dimensional space would I see there would this be one for level zero two for level one three for level two is that correct and then because level two would correspond to us that's not correct okay dimensional space spinners or no no I know I was saying that level one that's your spinner that's just spin 1/2 right so that would be 2 right and then level 2 that's your spin one object isn't it no I'm sorry no if you if you consider a spinner for a 3-dimensional space that is only a two component object so it Pascal's triangle one two one so when you get to the next level in three dimensions I'm sorry I said when you say two dimension do you mean three space show and or you mean three spatial or what general I know I I meant okay that's the uncontrol I'm confused I I was I was just thinking about ordinary so3 one ah okay dimensional space yes so the one four six four one that's a that's a Pascal's triangle wait a second so so your spinner in that case will be a Dirac spinor or it will be a my round a vial spinner with four components okay so one four six four one and that six would be composed into a 1 a 4 and a 1 and that the floor that you're looking for is there so take upwards Stephen if you look at what they even have on the screen in or dimensions it reads 1/4 at the next level that actually means 1 or +1 4 plus 1 you see how it level 4 in this diagram there are two representations yes if you're in 4 dimensions there would be three such representations occurring at level two okay and what do those correspond to does correspond to what do those correspond to they correspond to four Mayland by Linear's a better any cementery and that means that it's a singlet it is the axial vector and it's the gamma v hold on a second hold on hold on hold on so you're saying in so you're saying there's a there's a vector these are the gosh this remember reminds me weak interaction theory so okay so women back in the day they used to be the scalar vector tensor axial vector pseudo Steven yeah okay that that takes me back a long way okay so it in your in your version of this thank you I don't know which one have you changed it there for adventures or thank you yes okay so women so there is so those are the those are the things I just mentioned the scale of Exile in years that's right okay so they're in a different order here okay so the last one is the gamma 5 1 which is the pseudo scalar one the first one is this is the yeah I get it I get it that tum okay so hold on level level one is just okay so they're Pilon is alright and then you have a trial in here which I don't know about otherwise why you will only have oh I see I see those have to be that's a actually I'm a little confused here so the level two there you have what are those in terms of the the so that last one looks like it's a pseudoscalar the one with the gamma five yes right and the one before it looks like it's an axial vector there are five gamma a that's also correct okay so and the one before that what is that C sub a B so talk about Steven but in super space it is useful to introduce emotion and in metric or the offenders okay and then object C is that object and the way that you would have studied in the old days you would have said I could right sidebar sigh as as the scalar mm-hmm I would written side Gamma Phi Gamma Phi Gamma mu Sai as the axial vector yep and I would have written sidebar Gamma Phi sigh as the pseudoscalar yep and so those things that you see it level two are precisely the same objects with this slight sort of jazz because it's super symmetry okay okay fair enough okay the trilinear thing would I be familiar with that foot from from I don't that doesn't ring a bell at all because you don't you know you you you don't end up with in lagrangians you don't end up with side side side types exactly so but it turns out that under the wrist transformation property of the object that pointing to you don't have exactly the same lesson tree there is transformation properties as the object at level one and that's why it's a four okay okay okay so so the interpretation I mean I'm still is there a spacetime interpretation in terms of spins and things of these various numbers or is this a different sort of some kind of super property I mean these properties were known before supersymmetry so we can't blame supersymmetry for these things no but I mean it's there an interpretation because because okay looping back to the thing that I'm most you know focused on here which is this question of you know in our models how do we understand from limits of graphs and things the presence of things like spin and so I guess I'm I'm curious what is it the interpret I mean these are these internal quantum numbers in the super symmetric set up or are these supposed to be space-time quantum numbers these are space-time quantum numbers because it's supersymmetry the theta carries the representation of a spinner okay so these are not internal quantum numbers they're they're spaced I like numbers so so if I were to look at this particles that come out of this theory what spins would those particles have that's shown on on equation three point four okay the first particle which is called V zero is a scalar okay the second particle which is called V alpha one that's a spinner mm-hmm at the next level there are three different particle types there's another scalar which is V 1 2 there's a pseudo scalar which is V 2 to know the pseudo scalar V 2 2 and there's an axial vector which is V B 2 yeah okay but what about off to that so those ones I sort of understand those okay by axial vector we mean takes me back that means J to the P of 1 plus is correct ok so then what goes on off to that what supersymmetry you can't stop you have no power cities there's no justification for stopping the expansion and beta so you have to get you have to keep going and see okay so what's the next one what is what is the what is the spin of the next object it's been 1/2 the 3 theta thing has been 1/2 yes remember I Mitch transforms under rich transformation in exactly the same way at level four level three so it's got to be spend one hair I see so there aren't highest men so I mean I vaguely remember that that the you know just in the particle content of supersymmetric theories that there are a bunch of spree spin three-halves particles and things that come out all the way if you go to a different bolt of web I see okay so so here you've got an additional spin 1/2 particle and that thing with four Thetas is is what is that is that a spin zero again or what yes it is okay but that's not the end of the expansion I would go out to fly yesterday's experience that's because oh that's right because it's in three dimensions that's right the the number of but in 11 dimensions or whatever ah indeed that's what they were showing you with that thing boy from one to sixteen okay okay I'm finally now vaguely understanding okay so that was essentially the particle content of your thing in your N equals 11 case well I'm sorry that was too mister they were showing about 11 okay all right 10 dimensional okay so so then so what's happening can we go back to that that one I wanted to see the the 10 dimensional thing here um okay so are these just alternating between spin 1/2 and spa I don't know what spin 1/2 spin 0 spin 1 well yeah what what do you mean the analog and high dimensional heat so the spin I mean there's still a notion I mean this this relates again to this s OD for arbitrary D I mean do you consider for D equals 10 for example do you still think of that as you know I mean you're not changing there's still a very well-defined notion of spin 1/2 which is the spinor representation of s oh 10 right and yeah it's just that the number of components is is larger that's correct that is correct okay so your 20 here what is that at level 2 oh that's actually a ordinary okay okay so so I mean the graph theoretic interpretation of this is how do you how do you sort of think about it as sort of a metal level this is this is sort of well actually actually no I'm gonna ask a different question you know given this collection of numbers this is a very specific collection numbers yes and yes let me say then that the set of numbers you can just get by taking the waste product of 16 in all possible ways okay and then decomposing those representations into irreducible representations of so1 die that's what they come from okay but but so the thing that I'm curious about is is if you are - if you were given data about these irreducible representations and you're asked what group do these correspond to and let's imagine because you know the specific group s o1 9s o 1/8 et cetera they will have different collections and numbers all right but if I were to be very coarse and I would just to throw you a random collection of numbers yes you would say oh that's not a compactly group that doesn't correspond to the the right numbers to form a you know a nice whatever a compact superly group or whatever probably and so what I'm trying to understand is what would it what would happen in other words does there exist an object a mathematical object which given some more arbitrary collection of numbers is the thing that is the the sort of the reconstructed continuous legal-like thing that is approximated by C and what I'm getting at is what we're gonna get out of these models of ours is something which in some continuum limit might approximately group but we're not going to get at you know it's a discrete object and what I was really curious about with your Adinkra is if I were to already kind of you know pick it instead of a nice Dinka graph that comes from Ali group if I have to go the other way round right and say here's a family of progressive I don't know pieces of a tinkerers or something yeah I just you know smush them together and I say does this family limit to something right I understand your question Stephen I don't know the answer it's a great question because the session you're saying there's someone gives me some dado of multiplets and say dimensionalities mm-hmm how do I recover the groups that these things are parts of it I don't I don't even know how to think about that okay so I haven't ever thought him so and so you're I mean if um um actually let me could I could you unsure for a second I'll share for a second here hold on let me just um this is Steve I'm gonna I have a hard stop and about another 20 25 minutes or so okay okay no I I can go I don't know other people can stay or now Steve can I make a comment a little bit later yeah I was just gonna ask one one thing I'm curious about yes please my coma but I just I think I for for your question because but I don't want to hope held anyone here so I'll just it's not about a dinkus well though I get a better idea by that dinkus from this talk but it's not about it's about this spin that you're talking about I spin representations let's let's uh okay I have a guess here so this you know this here is obviously a four dimensional cube hypercube and we could you know we go to five dimensional whatever now imagine that I were to modify this a bit and I'm asking the question what is the the the not quite the four and a half dimensional hyper cube low because these are just graphs what I have here is just a graph and so you know I just have the connectivity data here so I mean I could you know if I take this and you know that's just a and so I guess what I would imagine is that the average coordination number here I mean this is exactly the way that we would compute the volume of JDC balls is if I change the average coordination number to not be equal to in this case you know what is it five or something but if that average coordination number is not five but is something different is D then I think that's the analog I mean so in other words if you ask what is the Adinkra diagram the corresponds to D equals you know seven and a half okay my claim would be that what it will be as a graph that um has average coordination number you know whatever seven a half or whatever anyway that was just my thought yeah I I agree that that's plausible for that can I ask two quick questions to Jim like related to that yeah okay sure so um okay the first question is very very simple minded it's just that so when the this construction of a tinkerer of the combinatorial construction you showed earlier so the best of my understanding this is a this is a special case for one-dimensional superly algebra yes right so then when you're doing for instance in in the when you're doing these things in the four dimensional the ten dimensional case is there some kind of clue to climb like reduction that you have to make in order to be able to do that do you think the adding crack representation yeah that's how a great question the answer is yes basically we we use a kind of holography but that will be first certainly when we first started this we only start with a four dimensional theory and then we say what is left of the algebraic structure if the feels only dependable of time right that's how do you think we emerge that makes sense okay so and does that work in arbitrary dimensions or is it specific to four and ten arbitrary integer images I've never tried for non integer fair enough okay and and then and then the and then the dimensionality of the Adinkra is just the number of supersymmetry generators that's career so it has nothing to do with the actual dimensionality of the Leal career okay okay so so relatedly I want to ask this because I think this was related to Stevens Point about coordination numbers I average coordination numbers how exactly do I haven't a ding Kragh how do I go about getting a corresponding superb Riemann surface ah well that's a great question and in the really weird thing is for a large class of four colored incres we know what the answer is but I'm not the expert on that Mike colleague Kevin diga if he's still online I'm sure he'll be happy to talk to you about that ok yeah oh yeah he does seem to be oh here's yes here ok yeah I'm finally on the right place yeah so this has to do with ok what what you need to do for an ad incra is remember all these different colors what you need to do is you need to order these colors cyclically so you have different you know let's say you have red green and blue you'd have to put them in an order cyclically and then you have to fill in the faces where there are made of cycles that have adjacent colors in that cyclic ordering okay so for example if there's you know I don't know we can bring up in a dinkar example let me unshare and somebody with it let me look for let me look for a paper that might serve this but so essentially what you're doing then is you're taking this graph and you're filling in faces and some words right so here's the problem if you if you fill in also so what what are the faces now in if you have an ad in crow usually what you'll well what you have my definition is you have these different colored edges and if we only focus on two colors at a time what you will end up with is many pull up a oh I know I'll just pull up my um trying to find a good picture here well I want to keep on talking about where I look for a good picture yeah so if you if you just draw um let's see in the Dinka okay let's see here's the book I'm writing yeah so here's in the Dinka and if you just look at the green and black edges for example you see that they form a quadrilateral a four cycle yep and hope this is supposed to be blue I don't know I got this color wrong but if this is why by the way this is why every diagram I drew is generated by program ah there we go oh I would never get any of them right go ahead please so if you fill in all of these with two cells like it's like just fill them square here the problem is in general you won't get a surface because you will have too many squares that share at the same edge for example so normally in a surface an edge it's it's an edge it's going to be bounded by two different phases but if you just fill in all of the two different pairs of two colors you'll end up with too many too many phases I changed Jason to each edge so the idea is you you pick an ordering of the colors in this case black green blue orange and then circle or inspector black and you only fill in the ones that have adjacent one so black green you fill in those green blue you fill in those blue orange you fill in those orange black you fill in those and if you do that well you'll end up with a surface and more than that actually you you end up with what is called well if the graph embedded in a surface and it you end up with what could be called the descent on fall and boy that's really weird my my daughter who's a math graduate student is actually actually happens to be studying those things for a completely different reason right now that's awesome okay yes fits into that whole story and these dashed edges the dashing actually is exactly what is needed to specify the spin structure on these surfaces in this case you actually get a torus in in other cases you'll get other other other kinds of surfaces and um then there's a whole bunch of work by SEMA so me Andres chicken that relates this to see super every month or is getting a a super structure like a super manifold structure on this which is really weird because you know this is motivated by supersymmetry but we didn't try to put supersymmetry into the geometry of this and yet supersymmetry comes out at the other end so we're still a little bit puzzled as to as to if that means something deep or if that's just a coincidence so so let me just understand so I mean you've got these add-in cards which are laughs and rain filling in faces so that you and then these become graphs embedded on surfaces basic that's correct and so what I mean and you're saying can you get all kinds of exotic geniuses and exotic forms of services from more more sophisticated interests sure yeah so I'm not sure what you mean by exotic I mean you we get a we get a bunch of different services yeah that's exactly right but I mean just just labeled by by betting numbers or something I mean they know oh yeah I mean yeah so we can we can of course calculate the genus just so you can do that pretty easily because we know how many vertices it has of course it depends on which is deeper you're looking at and each vertex has one edge of each color so depending on how many colors you have and of course each edge comes from two vertices so that tells you that similarly the faces are all going to be square faces so each vertex has n of those as long as you restrict it to the you know the cyclic permutation of colors and each face has four edges on it and so you can write out a formula for if you know the number of vertices the number of colors you can write a formula for the number of edges the number of faces do the Euler characteristic and that tells you the Dinah's so it turns out to be n minus 4 times 2 to the number of vertices I think there might be minus something something like that plus 1 so the only reason you you get a design door phone for a super Riemann surface rather than an ordinary agreement surface it's just because of the sign change induced by the dashing exactly right ok and but so the ordinary like climb classification of Riemann surfaces in terms of des on don't fall like places certain restrictions on like ramification type and things like that and those still hold yes so I guess there's another piece when saying to sound and I'm not talking about a surface only I'm talking about a surface together with a map from that surface down to three months to year and I know yes indeed that that does happen so oh there is a you know on this on the Riemann sphere you'll have a a single boson and a single firm neon and you will have well there I have to say they're they're sort of two different bases here that the bottom base just is just you know the standard you know you got a boson and a fermion connected by a single edge but above that you have an n-fold ramification ramified cover basically of Z to the N if Z is complex number great cover of this where you have one edge of each color connecting to two vertices so I don't have a picture of this here but you know basically imagine a beta beach ball you've got a sphere and you've got a vertex your vertex here you've got all these different colored edges going between them I see what you're saying yeah okay so then you have a cover this thing would be a cover of that right right have ramification points in the middle of each of those little wedges of the beach ball which presumably just and Jonathan your paper with this actually and Kevin people people on a live stream are asking what the title of your book is because they want to know when it comes out just have to look for your name so far but I don't know that may or may not change yeah Kevin just the thing you were saying earlier about covering you know interpreting these things as covers of the surface is really interesting so does that mean you can effectively associate a monitor meat group where they get across exactly Wow okay and what's the restriction on the monetary on the monitor right so so there actually there are two different versions there's a version that that's just ordinary monodromy group which actually turns out to be well okay so it's just Z 2 to some power but what's interesting is is that actually sits inside a bigger Z 2 to some power higher power and it's actually better to think of the monodromy group as a code a linear code ah yeah okay and the criteria in the for it to have this dashed you know have this criterion where each face has an odd number of - edges that is if and only if that code is doubly even so with a multiplicity for yes okay yeah okay that makes sense the monodromy if there's another version of monodromy is what happens if every time go on and ask edge you imagine that you are flipping between two different versions a positive and a negative so you have assigned monodromy group and and that is actually a Clifford Clifford group right right that covers the code so these are linear codes over f2 presumably first amazing case so actually that relates to the question we all write in the beginning about about power of statistics and things so you know you can only define a you know obviously a vector field over a field of a fuel FP where P being prime right does that not place restrictions on like on how you can grade things effectively you know it is there something special about the fact about Z - grading that comes from the fact that 2 is a prime yeah these sort of philosophical questions that are I mean if you had a Z 3 graded algebra then this correlation fail right because you wouldn't get a vice right so Z 2 is showing up on all sorts of different different places and they kind of into all interlace in this kind of fancy way so the Z 2 is showing up because of course you got and fermions right it's also showing up because you got solid edges and - ditches and those play around in some kind of nice way oh I should also mention um if you for those in the audience who know a little bit about differential topology I will say that there is an analogy and in fact it's the the surface actually makes it not just an analogy this actually makes it correct where the the choosing between solid and and you know the disposals and fermions making a part by partition is essentially like choosing an orientation and the solid and - edges is similar to choosing a spin structure and the corresponding thing is in differential topology you've got Stifel whitney classes W 1 W 2 and W 1 really controls whether or not you have an orientation and that has to do with whether or not for each of these things you know you have a edge between any two you know connects two different statistic particles and the can see for the seconds people would in class that governs the spin structures that has to do with the odd dashing sort of thing so from my point of view the z2 is really a matter of the fact that the characteristic classes for vector bundles really take values in Z MA - ok that's fascinating so that means that the the statement the gym made earlier about the fact that you can construct this bike partition is really just saying that when you embed the graph in a surface the surface has to be orientable that's correct that's that's that's a really cool correspondence yeah yeah ok well yeah thanks for that that's that yeah that clears things up ok so let's see I think Jim had to go very soon and unfortunate evening it's almost fun I agree it's fun we need to do again sometime yeah yeah right no I I didn't you can't see I mean young gray and hazel start to tell you a really cool story where we've been able to use ideas from a dangerous looking at the low energy limit of in theory and we're the first people to be able to to figure out the two billion 147 million four hundred eighty three thousand six hundred forty eight degrees of freedom how do they represent the Lorentz group and we got all kinds of cool thing waiting you know me that there are that number of different kinds of particles map states states that particles the states how many particles one so there are 1000 second there are 1000 and 1040 1040 at 94 bosons and one thousand eleven hundred eighty-six brimmy ons that's going to be exhilarated a lot of them are going to be exactly ok so there's not so much work for experimentalists to go and try and do but there's a ton of work for for theorists here I mean the problem that I was telling you about that I first started trying to solve with John's Schwartz is really is the same problem effectively that I'm trying to now work on younger II with younger II and hazel in the context of in theory is essentially the problem shows up in lots of different guises so there's I mean there's so much work to be done here mathematically it's ridiculous right I mean the thing look the the contact with the kinds of things we're thinking about which I don't fully understand but but you know we have this very simple underlying set up and we're trying to take essentially we're trying to understand how in the continuum limit certain kinds of things emerge your for example and I guess the question well I don't really so let me make a proposal Stephen yeah since I think the common feature in both of our investigations are at least from my point of view as graph theory yes and so what really might be useful to try to use our results where I mean look we got a lot of assumptions built in our result because we're only interest in supersymmetry but to somehow try to use our results from graph theory to see if they're as you said in a certain limiting case you can see perhaps i graph somehow coming out of your work yeah aspects of your graphs that's the thing that I think is most interesting here right well so right but I think the the sort of the contact point there would probably be to you know this thing about fractional dimensional I mean to me this thing about how do you generalize ordinary lis groups because you know you've you know you're using these graphs to somehow illuminate how representations of these groups work correct and what I think is going to happen is that your your graphs are nice sort of perfect cases where things close because they're they're you know integer dimensional space what I'm guessing is that there is a you know that if you're not as perfect as that that in some limit you will get you'll get something where in specific cases where it's you know nice interdimensional space you'll get what you have and then you'll get these weird intermediate cases where things don't close properly but where you can still say as some approximation that you're getting some something analogous to some let's say fractional dimensional Lee group yeah and then you know for us the then the question is can we get because what we will naturally have is things that are these messy unclosing kinds of things which you know might turn out to have limits that are correspond to - you know integer dimensional space or might not but I think that's the somehow that's the that's the point of contact is that you're representing things in you know the question is can we go from our graphs to something where we understand how to extract something that is like representations of some symmetries of our graphs and then come and come and go from what you're doing where you are going from understood lis groups so to speak to a graphical way of representing representations yes I mean that that's that will be the the the potential correspondents agree um yeah that's interesting I I think you know we had I know you have to go but I noticed that okay okay okay I mean we were we were do we still have questions about fractional derivatives or did we did we um well I just I just hope someone will actually do the calculation that's indicated by my suggestion and give me the answers I'm really curious as to what the answer looks like all right well I think we looks like we have Oleg who's even Harewood by a video actually Oleg and I have been talking and a chat in a sidebar okay okay fine fine all right well that yeah he's the he's the expert on this stuff um and um yeah and I'm I'm it looks like we're not gonna have a chance to see the end of the hazel and angry we are doing well we don't know yet because this is actually our very first one so we're going to be on I mean we and the question is how long can I stay away from my day job and work primarily yeah yeah right right right no I the answer is we're going to do a bunch of these and it'd be it will be it will be great to continue this at some point I mean we've been doing sort of live streams of some kind almost every day but many of them are are our sort of internal efforts to understand different kinds of things but I think the maybe some of the folks who are on here can can separately interact and see whether we can get to sort of the the the next stage of of understanding I mean again I'm sure there is some correspondence between you know I'm I just have this gut feeling that there's some way of using representations to sort of approximate I mean you know to somehow form this way of thinking about some approximate generalization by the way does anybody on this on the stream like I really understand that the lien categories fusion categories business is that um somebody here like a like really really know how that all works Kevin woodchuck be able to do that for us yeah I would guess Chuck would would be the person I would start by asking but I don't know that for a fact okay fair enough I mean I think because that's the that's the closest case that I've seen of anybody trying to define these fractional dimensional um Lee groups and I mean maybe see see the thing is I mean and then in terms of the supersymmetry aspect of it well we didn't really get a chance to talk about this but the whole what is a Fermi on what is a boson really right how do we try to understand that yeah and I mean for you that you know you've been living fermions and bosons and putting them in super super algebra and so on forever but you know for us we have a pretty definite idea of what um what they might correspond to it in our systems which which um but um yeah we could explain that it's important by the way it's worth mentioning that the the fusion category story is deeply related to the supersymmetry story as I'm sure many people here know the the tensor categories of representations of super groups are exactly the Deline categories and so at some level you can think of these these generalized tensor product spaces as being the ones that are spanned by by super multiplets so what so where does the fractional dimensional thing come in there well from like I was saying earlier from from the fact that when you do when you apply this generalized integral functor you can end up with Adaline tensor category where your tensor product basically has a non integer arity well well in good does that come out of the super multiplets I I don't understand supersymmetry well enough to know the answer I just know that there's this there's a Tanaka crying duality between the between the that sense of product categories and the and as I say these spaces spanned by super multiplets well so then the question would be for for these guys whether they're you know that that very long number that Jim was was reciting the question is if is there an intermediate case where it isn't that very long number but but that would only work if you didn't have you know if in the end your space correspondence is something that wasn't one of these nice integer dimensional things okay there are many questions we're not going to have a chance to address more of them here but that it's some this was great you you you've also done a very important meta thing which is you've proved that these kinds of discussions can actually work I make can I make a quick comment and I know I apologize for holding everybody was that be a sort of very similar like you you you would have a classical spin which is not quantized and then you're you know which is not integer but then you you want to somehow obtain an integer representations of that in our I've seen something like this if you you know you you imagine an angular momentum which is classical angular momentum and it's it can take any any values right but then in quantum mechanics spin only takes certainly so we have a guess I have a guess at least Jonathan doesn't yet agree with my guess but my guess is that in our in our systems you we think we know what angular momentum is and we think it has to do with if you take pairs of jd6 from a point that essentially define a plane you can then in a graph of any any hyper graph you can effectively define a generalized plane with pairs of gd6 emanating from a single point then you can essentially think of angular momentum as being sort of the the thing that you get by the you know in terms of edges that span across this plane in some sense and then the my so you can think about if you look at these edges you can kind of visualize them as something like a vortex of edges in some approximation it is my guess that there's a relationship between a kind of homotopy associated with the way those edges have to work and the way that one can and and the existence of essentially quantized spins oh yeah in physics they're app works that consider how how do you know it's a how you recover spin from semi-classical in the semi-classical limit see you know what which which part of that I mean that the semi-classical limit you're looking at very high spins right so you you and then how you if you want to reduce it to a quantized to a so it's it's in that I don't so I would have to collect myself and get get a little more but it's some kind of I've seen things like you go to where's the minute terms and then you you you recover the spin 1/2 or spin I think normal thinking about semi classical limits is that you're really looking at high spins I mean that's because the whereas what I'm thinking is that the way that essentially it's like you know you have a vortex in some vector field or something and you're asking there's there's a discrete you know there's a you know sort of topological it's a topological discreteness that comes about because in that particular case you might just have a you know some PI one of of your of your space and I'm you know this is very vague because I don't really know how it works yet I still understand quantization there's something you what's that it's like a quantization condition I could directly look at the rack string there's something you can yes that's right like it like a Dirac string but it but it would be a well go ahead Jonathan was about to well I'm just gonna say that the thing that still doesn't convince me about this is why if you're defiant if you're defining quantization of spin in terms of effectively topological obstructions in the causal network why you don't get quantization of classical angular momentum because you're not you're not using any features that are peculiar to the multi-way causal network as opposed to the spacetime causal network okay so that's a more complicated issue I I'm sorry - thank you thanks for joining us and we should probably um probably wrap this up now and continue another time okay well all I can say Steve is I'm the first time I tried to get you interested in your house was it what ten years ago five I can't remember we come a long distance if this discussion continues yeah yeah right right no no no you seem it but this is what I mean I've been I was off physics for quite a while and in case you didn't notice and I'm now back on physics so but also also I'm now this was very helpful in terms of understanding what you guys had been doing and I think you know I I am convinced that there's a you know we're operating from two ends of a bridge across a pretty you know across this this fast-moving River so to speak and I think yeah I mean it's it's not it's not it's non-trivial to fill in the the you know the middle of the bridge but I think it's a really interesting all right but we should we should wrap it up here and thanks thanks to everybody who's joined this some this zoom session and thanks to folks on a live stream and to be continued and thank you Jim okay very good Steven okay I have the conversation and goodbye to everyone thank you for joining us let's do this again sometime was fun great goodbye

Channel: Wolfram

Views: 16,846

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Keywords: Wolfram, Physics, Wolfram Physics, Wolfram Physics Project, Stephen Wolfram, Science, Technology, Wolfram Language, Mathematica, Programming, Engineering, Math, Mathematics, Nature, A New Kind of Science, NKS, Computer Science, Philosophy

Id: go9XxI1Igsk

Length: 161min 48sec (9708 seconds)

Published: Mon May 04 2020